Q29 of 39 Page 1

Prove that in a right-angle triangle the square of the hypotenuse is equal the sum of squares of the other two sides.


Draw a BD AC


In ∆ ADB and ∆ ABC


∠ BAD = ∠ BAC (same angle)


∠ ADB = ∠ ABC (90⁰)


By the AA criterion for triangle, similarity which states that if two triangles have two pairs of congruent angles, then the triangles are similar.


∆ ADB ≈ ∆ ABC



⇒ AB2 = AC×AD …(1)


In ∆ BDC and ∆ ABC


∠ DCB = ∠ ACB (same angle)


BDC = ABC (90⁰)


By AA theorem


∆ BDC ≈ ∆ ABC



⇒ BC2 = AC×CD …(2)


On adding eq(1) and eq(2)


AC×AD + AC×CD = AB2 + BC2


AC(AD + CD) = AB2 + BC2


AC×AC = AB2 + BC2


AC2 = AB2 + BC2


Hence proved.

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